THEOREM XVIII. 545. Similar triangles are to one another as the squares on their corresponding sides. Let the similar triangles ABC, AHK, be placed so as to have the sides AB, AC, along the corresponding sides AH, AK, and therefore BC || HK. Join CH. Since, by 505, triangles of equal altitude are as their bases, A ABC : A AHC :: AB : AH, and A AHC : A AHK :: AC : AK = AB : AH, by hypothesis ; EXERCISES. 92. The ratio of the surfaces of two similar triangles is the square of the ratio of similitude of the triangles. 93. If the bisector of an angle of a triangle also bisect a side, the triangle is isosceles. 94. In every quadrilateral which cannot be inscribed in a circle, the rectangle contained by the diagonals is less than the sum of the two rectangles contained by the opposite sides. 95. The rectangles contained by any two sides of triangles inscribed in equal circles are proportional to the perpendiculars on the third sides. 96. Squares are to one another in the duplicate ratio of their sides. THEOREM XIX. 546. Similar polygons are to each other as the squares on their corresponding sides. B Ć HYPOTHESIS. ABCD and A'B'C'D two similar polygons, of which BC and B'C' are corresponding sides. CONCLUSION. ABCD : A'B'C'D :: ABP : A'B'2. By 545, any pair of corresponding triangles are as the squares on corresponding sides, Δ ABD BD2 ABCD BC2 A A'B'D B'D2 A B'C'D therefore, by 496, A ABD + ABCD B C2 .. B'Cuz; In the same way for a third pair of similar triangles, etc. 547. COROLLARY. If three sects form a proportion, a polygon on the first is to a similar polygon similarly described on the second as the first sect is to the third. THEOREM XX. 548. The perimeters of any two regular polygons of the same number of sides have the same ratio as the radii of their circumscribed circles. PROOF. The angles of two regular polygons of the same number of sides are all equal, and the ratio between any pair of sides is the same, therefore the polygons are similar. But lines drawn from the center to the extremities of any pair of sides are radii of the circumscribed circles, and make similar triangles, PROBLEM II. 549. To find a mean proportional between two given sects. Let AB, BC, be the two given sects. Place AB, BC, in the same line, and on AC describe the semicircle ADC. From B draw BD perpendicular to AC. BD is the mean proportional. (383. An angle inscribed in a semicircle is a right angle.) (521. If from the right angle a perpendicular be drawn to the hypothenuse, it will be a mean proportional between the segments of the hypothenuse.) EXERCISES. 97. If the given sects were AC and BC, placed as in the above figure, how would you find a mean proportional between them? 98. Half the sum of two sects is greater than the mean proportional between them. 99. The rectangle contained by two sects is a mean proportional between their squares. 100. The sum of perpendiculars drawn from any point within an equilateral triangle to the three sides equals its altitude. 101. The bisector of an angle of a triangle divides the triangle into two others, which are proportional to the sides of the bisected angle. 102. Lines which trisect a side of a triangle do not trisect the opposite angle. 103. In the above figure, if AF I AD meet DB produced at F, then A ABD = A FCB. PROBLEM III. 550. On a given sect to describe a polygon similar to a given polygon. Let ABCDE be the given polygon, and A'B' the given sect. * A'B'C' = X ABC, * B' A'C' = { BAC, .: X B'C'A' = BCA, and A A'B'C' ~ A ABC. Then make X AC'D = X ACD, * C'AD = % CAD, :: C'DA = CDA, and A A' C'D' ~ A ACD, etc. Therefore, from the first pair of similar triangles, AB B'C' :: AB : BC, and B'C' : C'A' :: BC: CA. From the second pair of similar triangles, C'A : C'D :: CA : CD, .:. B'C' : C'D :: BC : CD, and so on. Thus all the Xs in one polygon are = the corresponding s in the other, and the sides about the corresponding s are proportional. :. the polygons are similar. |